complete the table of values for the function: 1 x-2-1.5-1/2-.01-.0010 f(x) x21.51½.01.0010 f(x)
TRANSCRIPT
Complete the table of values for the function:
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x -2 -1.5 -1 -1/2 -.01 -.001
0
f(x)
x 2 1.5 1 ½ .01 .001 0
f(x)
Pre-Cal
Rational Functions And Asymptotes
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Rational Functions are not always continuous…they can have: Vertical asymptotes Horizontal asymptotes Holes Slant Asymptotes
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Asymptote: a line that a graph approaches but never touches
Hole: a point in which the graph is undefined but there is not an asymptote
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Vertical Asymptote: If y goes to +∞ or -∞ as x-values get closer and closer to some specific x-value, c, the x = c is a vertical asymptote.
Horizontal Asymptote: the y-value that f(x) is approaching as x goes to +∞ or -∞.
Slant (Oblique) Asymptote: a line that f(x) is approaching as x goes to +∞ or -∞.
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Simplify function first!
Vertical Asymptotes: Any points where the domain is restricted in the simplified function.
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x = 3 is the VA
x = -4 is the VA
x = -3 is the VA
Simplify function first!If any factor simplifies from the denominator then that x value becomes a hole
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x + 2 simplified so x = -2 is a hole. To find the y value plug -2 into the simplified function for x so y = 3/5. Hole: (-2, 3/5)
Simplify!Simplified value is x value of hole.
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x + 3 simplified so x = -3 is a hole. To find the y value plug -3 into the simplified function for x…so y = 6/7. Hole: (-3, 6/7)
Given
Simplify function first!!
If n < d, then y = 0 is the HA
If n = d, then is the HA
If n > d, then there is no HA (check for Slant asymptote)
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Simplify!n < d y = 0n = dn > d No HA
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HA: y = 0
HA: y = 2/1 = 2
HA: none
Simplify!n < d y = 0n = dn > d No HA
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HA: y = 0
HA: y = 3/1 = 3
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A Rational Function will have a SA if the Numerator has a degree of ONE more than the denominator
Simplify First!!
To find the SA divide the N by D using long division
The SA = the polynomial part of the quotient (the answer without the remainder)
A graph can pass through the slant asymptote.
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SA: y = x+ 5
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SA: y = -3x – 3
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SA: y = -2x – 4
X intercept: let y = 0
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y intercept: let x = 0
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VA: x = -1HA: y = 1Hole: (-3, 0)SA: NoneY int: (0,3)X int: None (it is a hole)
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VA: x = 2HA: y = 0Hole: NoneSA: NoneY int: (0,3/8)X int: None
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VA: x = 2HA: NoneHole: NoneSA: y = 3x + 8Y int: (0,-1/2)X int: None (imaginary zeros)
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VA: x = 1HA: NoneHole: NoneSA: y = 5x + 2Y int: (0,-1)X int: None (imaginary zeros)
Find the domain, all asymptotes, holes, intercepts, and zeros.
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VA: x = 3/2HA: y = 1/2Hole: (-1, 0)SA: NoneY int: (0,-1/3)X int: None (it is a hole)
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VA: x = -1HA: y = 1Hole: (-3, 0)SA: NoneY int: (0,3)X int: None (it is a hole)
Find all the asymptotes, holes, and intercepts.
Make a table of values.
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VA: x = 1 HA: y = 3 Holes: None SA: None Y int: (0,1) X int: (1/3,0)
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VA: x=3, x=-3 HA: y = 1 Holes: None SA: None Y int: (0,0) X int: (0,0)
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VA: x=-1 HA: None Holes: None SA: y=x – 2 Y int: (0,0) X int: (0,0), (0, 1)
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VA: x=3, x=-2 HA: y = 1 Holes: None SA: None Y int: (0,1/3) X int: (-1,0); (2, 0)
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VA: x=3 HA: y = 1 Holes: (-2, 3/5) SA: None Y int: (0,1/3) X int: (1,0)
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VA: None HA: None Holes: (5, 1) SA: None Y int: (0, -4) X int: (4,0)
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VA: x=1 HA: y = 1 Holes: (2, 3) SA: None Y int: (0,-1) X int: (-1,0)
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VA: x = 2 HA: None Holes: None SA: y = 2x – 1 Y int: (0,-5/2) X int: None
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VA: x=-1 HA: y = 1 Holes: (3, 3/4) SA: None Y int: (0,0) X int: (0,0)
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VA: x=-2 HA: y = 2 Holes: (1, 5/3) SA: None Y int: (0,3/2) X int: (-3/2,0)
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VA: None HA: None SA: None Holes: (-2, -1) Y int: (0,3) X int: (-3/2,0)
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B
A rectangular page is designed to contain 48 square inches of print. The margins on each side of the page are each 1½ inches. The margins at the top and bottom are each 1 inch. What should the dimensions of the page be so that the minimum amount of paper is used?
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A rectangular page is designed to to contain 57 square inches of print. The margins on each side, as well as the top and bottom, are all 1 inch deep. What should the dimensions of the page be so the the least amount of paper is used?
48